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kolbaska11 [484]
2 years ago
5

The photography club decides to publish a calendar to raise money. The initial cost is 600$. In addition to the initial cost eac

h calendar cost 2.50 to produce. The club plans to sell the calendars for 8$ each. Write and solve an inequality to find the number of calendars that the photography club must sell in order to raise at least 1200$
Mathematics
1 answer:
velikii [3]2 years ago
4 0

Answer:

110 calendars to have a profit of atleast 1200.

Step-by-step explanation:

Initial expense = 600

Let the number of calendars that the photography club must sell in order to raise at least 1200$ be x.

Cost per calendar = 2.5

Cost for x calendars = 2.5 x

Total expenses = 2.5 x + 600

Income per calendar = 8

Income = 8 x

Minimum profit = 1200

Therefore the inequality is,

8x - 2.5 x + 600 ≥ 1200

Solving we get,

5.5x ≥ 600

x ≥ 109.09

x ≥ 110

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